Give IUPAC name of the compound given below.
Question: Give IUPAC name of the compound given below. (i) 2-Chloro-5-hydroxyhexane (ii) 2-Hydroxy-5-chlorohexane (iii) 5-Chlorohexan-2-ol (iv) 2-Chlorohexan-5-ol Solution: Option (iii) 5-Chlorohexan-2-ol is the answer....
Read More →Which of the following compounds
Question: Which of the following compounds is aromatic alcohol? (i) A, B, C, D (ii) A, D (iii) B, C (iv) A Solution: Option (iii)B, C is the answer....
Read More →Mark (√) against the correct answer in the following:
Question: Mark () against the correct answer in the following: $f: R \rightarrow R: f(x)=x^{3}$ is A. one - one and onto B. one - one and into C. many - one and onto D. many - one and into Solution: $f(x)=x^{3}$ Since the function $f(x)$ is monotonically increasing from the domain $R \rightarrow R$ $\therefore f(x)$ is one -one Range of $f(x)=(-\infty, \infty) \neq R$ (codomain) $\therefore f(x)$ is into $\therefore f: R \rightarrow R: f(x)=x^{3}$ is one - one into....
Read More →The process of converting alkyl halides into
Question: The process of converting alkyl halides into alcohols involves_____________. (i) addition reaction (ii) substitution reaction (iii) dehydrohalogenation reaction (iv) rearrangement reaction Solution: Option (ii)substitution reactionis the answer....
Read More →CH3CH2OH can be converted into
Question: CH3CH2OH can be converted into CH3CHO by ______________. (i) catalytic hydrogenation (ii) treatment with LiAlH4 (iii) treatment with pyridinium chlorochromate (iv) treatment with KMnO4 Solution: Option (iii)treatment with pyridinium chlorochromate is the answer....
Read More →What is the correct order of reactivity of alcohols
Question: What is the correct order of reactivity of alcohols in the following reaction? ROH + HCl (ZnCl2) RCl + H2O (i) 1 2 3 (ii) 1 2 3 (iii) 3 2 1 (iv) 3 1 2 Solution: Option (iii)3 2 1is the answer....
Read More →How many alcohols with molecular formula
Question: How many alcohols with molecular formula C4H10O are chiral? (i) 1 (ii) 2 (iii) 3 (iv) 4 Solution: Option (i) is the answer....
Read More →Monochlorination of toluene in sunlight
Question: Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields. (i) o-Cresol (ii) m-Cresol (iii) 2, 4-Dihydroxytoluene (iv) Benzyl alcohol Solution: Option (iv)Benzyl alcohol is the answer....
Read More →Mark (√) against the correct answer in the following:
Question: Mark () against the correct answer in the following: $f: R \rightarrow R: f(x)=x^{2}$ is A. one - one and onto B. one - one and into C. many - one and onto D. many - one and into Solution: $f(x)=x^{2}$ $\Rightarrow y=x^{2}$ in this range the lines cut the curve in 2 equal valued points of $y$, therefore, the function $f(x)=x^{2}$ is many one. Range of $f(x)=(0, \infty) \neq R$ (codomain) $\therefore f(x)$ is into $\therefore f: R \rightarrow R: f(x)=x^{2}$ is many - one into...
Read More →Evaluate the following integrals -
Question: Evaluate the following integrals - $\int(x+2) \sqrt{x^{2}+x+1} d x$ Solution: Let $I=\int(x+2) \sqrt{x^{2}+x+1} d x$ Let us assume $x+2=\lambda \frac{d}{d x}\left(x^{2}+x+1\right)+\mu$ $\Rightarrow x+2=\lambda\left[\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(x)+\frac{d}{d x}(1)\right]+\mu$ We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$ and derivative of a constant is 0 . $\Rightarrow x+2=\lambda\left(2 x^{2-1}+1+0\right)+\mu$ $...
Read More →Mark (√) against the correct answer in the following:
Question: Mark () against the correct answer in the following: $f: N \rightarrow N: f(x)=x^{2}+x+1$ is A. one - one and onto B. one - one and into C. many - one and onto D. many - one and into Solution: In the given range of $N f(x)$ is monotonically increasing. $\therefore f(x)=x^{2}+x+1$ is one one. But Range of $f(n)=[0.75, \infty) \neq N$ (codomain) Hence, $f(x)$ is not onto. Hence, the function $f: N \rightarrow N: f(x)=\left(x^{2}+x+1\right)$ is one - one but not onto. i.e. into...
Read More →Classify the following compounds as primary,
Question: Classify the following compounds as primary, secondary and tertiary halides. (i) 1-Bromobut-2-ene (ii) 4-Bromopent-2-ene (iii) 2-Bromo-2-methylpropane Solution: (i) 1-Bromobut-2-ene is a primary halide (ii) In 4-Bromopent-2-ene, the bromine is attached to the secondary carbon atom, hence this a secondary halide. (iii) In 2-Bromo-2-methylpropane, the bromine is attached to the tertiary carbon atom, making it a tertiary halide....
Read More →Why is the solubility of haloalkanes
Question: Why is the solubility of haloalkanes in water very low? Solution: Haloalkanes are very slightly soluble in water because, to dissolve a haloalkane in water, energy is required to overcome the attractions between the haloalkane molecules and also to break the hydrogen bonds between water molecules....
Read More →Which of the products will be a major product
Question: Which of the products will be a major product in the reaction given below? Explain. CH3CH=CH2 + HI CH3CH2CH2I + CH3CHICH30 (A) (B) Solution: The molecule (B) will be the major product in the reaction. This addition reaction is carried out by following Markovnikoffs rule, wherein a double bond, the hydrogen from the hydrogen halide is added to the carbon atom with the most hydrogen atoms attached to it,...
Read More →Mark (√) against the correct answer in the following:
Question: Mark () against the correct answer in the following: $f: N \rightarrow N: f(x)=2 x$ is A. one - one and onto B. one - one and into C. many - one and onto D. many - one and into Solution: $f(x)=2 x$ For One - One $f\left(x_{1}\right)=2 x_{1}$ $f\left(x_{2}\right)=2 x_{2}$ put $f\left(x_{1}\right)=f\left(x_{2}\right)$ we get $2 x_{1}=2 x_{2}$ Hence, if $f\left(x_{1}\right)=f\left(x_{2}\right), x_{1}=x_{2}$ Function $\mathrm{f}$ is one - one For Onto $f(x)=2 x$ let $f(x)=y$, such that $y ...
Read More →Which of the following compounds (a)
Question: Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4. Explain why? Solution: A mixture of NaBr and H2SO4 gives Br2 gas as a product. Molecule (b) will not react with Br2 gas because of the stable molecule that is formed due to resonance stabilization....
Read More →Discuss the role of Lewis acids in the preparation
Question: Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark. Solution: Aryl bromides and chlorides can be prepared from arenes by electrophilic substitution. This reaction is carried out by treating the arene with chlorine or bromine in the presence of iron (III) chloride in the absence of light. Iron (III) chloride, is a Lewis acid, which generates the electrophile required to take the reaction forward. FeCl3 forms a coordination compound with Cl2, ma...
Read More →Haloarenes are less reactive than
Question: Haloarenes are less reactive than haloalkanes and haloalkenes. Explain. Solution: This is because of the resonance stabilization of the aryl ring. If we take an example of C6H5-Cl, there will be a conjugation of chlorine electrons with the electrons in the ring. This resonance makes C-Cl bond to get partial double bond character and makes less reactive to nucleophilic substitution....
Read More →Why iodoform has appreciable antiseptic
Question: Why iodoform has appreciable antiseptic property? Solution: Due to the liberation of free iodine iodoform has an appreciable antiseptic property....
Read More →Which of the compounds will react faster
Question: Which of the compounds will react faster in SN1 reaction with the OH ion? CH3 CH2 Cl or C6H5 CH2 Cl Solution: C6H5 CH2 Cl will react faster in an SN1 reaction with the OH- ion. This happens due to the stability of the carbocation in the compound. C6H5 group is already stable due to resonance, and the CH2 attached will gain that stability, thus forming a stable C6H5CH2+ carbocation after the cleavage in the first step of the SN1 reaction....
Read More →Evaluate the following integrals -
Question: Evaluate the following integrals - $\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$ Solution: Let $I=\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$ Let us assume $2 x-5=\lambda \frac{d}{d x}\left(2+3 x-x^{2}\right)+\mu$ $\Rightarrow 2 x-5=\lambda\left[\frac{d}{d x}(2)+\frac{d}{d x}(3 x)-\frac{d}{d x}\left(x^{2}\right)\right]+\mu$ $\Rightarrow 2 x-5=\lambda\left[\frac{d}{d x}(2)+3 \frac{d}{d x}(x)-\frac{d}{d x}\left(x^{2}\right)\right]+\mu$ We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\...
Read More →Let f and g be two functions from R into R
Question: Let $f$ and $g$ be two functions from $R$ into $R$, defined by $f(x)=|x|+x$ and $g(x)=|x|-x$ for all $x \in R .$ Find $f \circ g$ and $g \circ f$. Solution: To Find: Inverse of f o g and g o f Given: $f(x)=|x|+x$ and $g(x)=|x|-x$ for all $x \in R$ fog $(x)=f(g(x))=|g(x)|+g(x)=|| x|-x|+|x|-x$ Case 1$)$ when $x \geq 0$ $f(g(x))=0$ (i.e. $|x|-x)$ Case 2$)$ when $x0$ $f(g(x))=-4 x$ $g \circ f(x)=g(f(x))=|f(x)|-f(x)=|| x|+x|-|x|-x$ Case 1 ) when $x \geq 0$ $g(f(x))=0($ i.e. $|x|-x)$ Case 2$...
Read More →Aryl chlorides and bromides can be easily prepared
Question: Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does the preparation of aryl iodides requires the presence of an oxidising agent? Solution: Iodination of arenes is reversible due to the formation of HI. To move the reaction forward, an oxidizing agent like HNO3 or HIO4 oxidises HI, thus stabilizing the product....
Read More →Evaluate the following integrals -
Question: Evaluate the following integrals - $\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$ Solution: Let $I=\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$ Let us assume $2 x-5=\lambda \frac{d}{d x}\left(2+3 x-x^{2}\right)+\mu$ $\Rightarrow 2 x-5=\lambda\left[\frac{d}{d x}(2)+\frac{d}{d x}(3 x)-\frac{d}{d x}\left(x^{2}\right)\right]+\mu$ $\Rightarrow 2 x-5=\lambda\left[\frac{d}{d x}(2)+3 \frac{d}{d x}(x)-\frac{d}{d x}\left(x^{2}\right)\right]+\mu$ We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\...
Read More →Prove that
Question: Let $A=R-\{2\}$ and $B=R-\{1\} .$ If $f: A \rightarrow B: f(x)=\frac{x-1}{x-2}$, show that $f$ is one-one and onto. Hence, find $f^{-}$ 1. Solution: To Show: that $\mathrm{f}$ is one-one and onto To Find: Inverse of $f$ [NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)] one-one function: A function $f: A \rightarrow B$ is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for $...
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