Chemistry NEET Questions — Free MCQ Practice

Chemistry MCQs for NEET — Practice Questions with Answers

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Compounds I and II can be distinguished by using reagent

(I)	(II)
4–Amino–2–methlbut–3–en–2–ol	4–Amino–2, 2–dimethylbut–3–yn–1–ol

NaNO₂/HCI

Br₂/H₂O

HCI/ZnCI₂ (anhydrous)

Cu2CI₂ + NH₄OH

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Which of the following statements is true?

CH₃CH₂S^- is both a stronger base and more nucleophilic than CH₃CH₂O^-

CH₃CH₂S^- is a stronger base but is less nucleophilic than CH₃CH₂O^-

CH₃CH₂S^- is a weaker base but is more nucleophilic than CH₃CH₂O^-

CH₃CH₂S^- is both a weaker base and less nucleophilic than CH₃CH₂O^-

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Explanation

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$\begin{matrix} \underset{|}{{CO}_{2} H} \\ {CO}_{2} H \end{matrix} + \begin{matrix} \underset{|}{{CH}_{2}} & - & \overset{18}{OH} \\ {CH}_{2} & - & \overset{18}{OH} \end{matrix} \to_{∆}^{H^{+}}$ (A); Product (A) is:

$\begin{matrix} \overset{O}{\overset{∥}{\underset{|}{C}}} & - & O & - & {CH}_{2} \\ \underset{O}{\underset{∥}{C}} & - & O & - & {CH}_{2} \end{matrix}$

$\begin{matrix} \overset{O}{\overset{∥}{\underset{|}{C}}} & - & \overset{18}{O} & - & \underset{|}{{CH}_{2}} \\ \underset{O}{\underset{∥}{C}} & - & \underset{18}{O} & - & {CH}_{2} \end{matrix}$

$\begin{matrix} \overset{^{18} O}{\overset{∥}{\underset{|}{C}}} & - & O & - & \underset{|}{{CH}_{2}} \\ \underset{^{18} O}{\underset{∥}{C}} & - & O & - & {CH}_{2} \end{matrix}$

$\begin{matrix} H & - & O & - & \overset{O}{\overset{∥}{C}} & - & \overset{O}{\overset{∥}{C}} & - O & - & {CH}_{2} & - & {CH}_{2} & - & OH \end{matrix}$

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Explanation

The reaction involves the electrophilic addition of H+ to the carbonyl group of the carboxylic acid, followed by the nucleophilic addition of H2O18 (labeled water) to the resulting carbocation. The product will have the 18O label incorporated into the carboxyl group.

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Which are not cleaved by HIO₄?
I : glycerol
II : glycol
III: 1, 3-propanediol
IV: methoxy-2-propanol

I, II, III, IV

I, II

II, III

III, IV

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Explanation

1,2- or vicinal diols are cleaved by periodic acid, HIO₄, into two carbonyl compounds.
The reaction is selective for 1,2-diols.
The reaction occurs via the formation of a cyclic periodate ester

HIO₄will not oxidise, diol from 1, 3 atom & not used for cleavage of ether.

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Succinic acid $\overset{∆}{\to}$ (A) $\to_{∆}^{{NH}_{3}}$ (B) $\to_{KOH}^{{Br}_{2}}$ (C); Product (C) will be:

$\begin{matrix} \underset{|}{{CH}_{2}} & - & {CO}_{2} H \\ {CH}_{2} & - & \begin{matrix} {CH}_{2} & \begin{matrix} - & {NH}_{2} \end{matrix} \end{matrix} \end{matrix}$

$\begin{matrix} \underset{|}{{CH}_{2}} & - & {CO}_{2} H \\ {CH}_{2} & - & {NH}_{2} \end{matrix}$

$\begin{matrix} \underset{|}{{CH}_{2}} & - & {CO}^{-}_{2} K^{+} \\ {CH}_{2} & - & {NH}_{2} \end{matrix}$

$\begin{matrix} \underset{|}{{CH}_{2}} & - & {CO}_{2} H \\ {CH}_{2} & - & \begin{matrix} {CH}_{2} & \begin{matrix} - & Br \end{matrix} \end{matrix} \end{matrix}$

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Explanation

The reaction involves the conversion of succinic acid to its potassium succinate salt by treatment with potassium hydroxide (KOH). The carboxyl groups are deprotonated, and the resulting negative charges are balanced by the potassium cations (K+).

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(A) _{→ B + C; (B) and (C) both gives +ve iodoform test.Compound (A) is:-}
C₅H₁₀O H₃O+

CH₃-CH=CH-O-CH₂-CH₃

H
l

CH₃-C-O-CH₂-CH₃
l
CH₃

CH₃-C-O-CH₂-CH₃
ll
CH₂

Both (b) and (c)

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A solution of Ph₃CCO₂ in conc. H₂SO₄ gives (X) when poured into methanol X is :-
O
ll

Ph₃C-C-O-CH₃

O
ll

Ph₂CH-C-O-CH₃

Ph3C-OCH3

Ph₃C-CH₃

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Explanation

When triphenylacetic acid (Ph3CCO2H) reacts with concentrated H2SO4 and the resulting solution is poured into methanol, the expected product is triphenylmethyl methyl ether (Ph3C-OCH3), which is option o3.

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0.092 g of a compound with the molecular formula C₃H₈O reaction with an excess of CH₃MgI gives 67.00 mL of methane at STP. The number of active hydrogen atoms present in a molecule of the compound is:

one

two

three

four

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Explanation

1 g of a compound react with excess of CH₃Mgl to give $\frac{x \times 22400}{92}$

0.092 of a compound react with excess of CH₃Mgl tp give $\frac{x \times 22400}{92} \times 0.092$ $\frac{x \times 22400 0.092}{92} = 67 mL of {CH}_{4} at STP x = \frac{67 \times 92}{22400 \times 0.092} = \frac{67 \times 92 \times 1000}{22400 \times 92} = \frac{670}{224} = 2.99 \approx 3 x \Rightarrow number of active hydrogen in compound . active H \Rightarrow 3$

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Optically active 2-octanol rapidly loses its optical activity when exposed to :

dilute acid

dilute base

light

humidity

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Test to differentiate between (CH₃OH) and (Ph-OH) is/are:

Litmus test

FeCl₃

Br₂/H₂O

All of these

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