Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C_8H_8O.Q undergoes Canninzzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction The option(s) with suitable combination of P and R, respectively, is (are)

Aldehydes Ketones and Carboxylic Acids PYQ — JEE Advanced 2017 · Shift 2 · Q5 with Solution

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Correct answer: A, B

Identify the nature of $Q$ and $S$ from the given tests

Both $Q$ and $S$ have molecular formula $C_8H_8O$ .

For degree of unsaturation:

\text{DU} = \frac{2C+2-H}{2} = \frac{2(8)+2-8}{2} = \frac{10}{2} = 5

So each compound has total unsaturation $=5$ , which strongly suggests an aromatic ring $(4)$ plus one more unsaturation, most likely a carbonyl.

Thus $Q$ and $S$ are most likely aromatic carbonyl compounds with formula $C_8H_8O$ .

Use Cannizzaro and haloform tests

For $Q$ :

$Q$ undergoes Cannizzaro reaction $\Rightarrow$ it must be an aldehyde without $\alpha$ -hydrogen.
$Q$ does not undergo haloform reaction $\Rightarrow$ it is not a methyl ketone and not acetaldehyde-type.

Among aromatic carbonyl compounds with formula $C_8H_8O$ , the suitable aldehyde is:

C_6H_5CH_2CHO \quad \text{(phenylacetaldehyde)}

but this has $\alpha$ -hydrogen, so it does not give Cannizzaro.

Another possibility is:

CH_3C_6H_4CHO \quad \text{(tolualdehyde)}

This is an aromatic aldehyde where the carbonyl carbon is directly attached to the ring, so there is no $\alpha$ -hydrogen on the carbon adjacent to the carbonyl. Hence it gives Cannizzaro reaction.

Also it is not a methyl ketone, so it does not give haloform reaction.

Therefore,

Q = CH_3C_6H_4CHO \quad \text{(methyl benzaldehyde / tolualdehyde)}

For $S$ :

$S$ undergoes haloform reaction $\Rightarrow$ it must contain the group

CH_3CO-

$S$ does not undergo Cannizzaro reaction $\Rightarrow$ it is not a non-enolizable aldehyde; a ketone is fine.

With formula $C_8H_8O$ , the aromatic methyl ketone is:

C_6H_5COCH_3 \quad \text{(acetophenone)}

This gives haloform reaction and does not give Cannizzaro.

Therefore,

S = C_6H_5COCH_3 \quad \text{(acetophenone)}

Now infer alkenes $P$ and $R$ whose ozonolysis give $Q$ and $S$ respectively

Ozonolysis cleaves a $C=C$ bond to form carbonyl compounds.

For $Q = CH_3C_6H_4CHO$ :

To get an aldehyde of type Ar-CHO by ozonolysis, the alkene should be of styrene type:

CH_3C_6H_4CH=CH_2

On ozonolysis:

CH_3C_6H_4CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3C_6H_4CHO + HCHO

So a methyl styrene (tolyl ethene) can be $P$ .

For $S = C_6H_5COCH_3$ :

To get acetophenone by ozonolysis, the alkene should be:

C_6H_5C(CH_3)=CH_2

On ozonolysis:

C_6H_5C(CH_3)=CH_2 \xrightarrow[Zn/H_2O]{O_3} C_6H_5COCH_3 + HCHO

So $\alpha$ -methyl styrene can be $R$ .

Conclusion

Thus the correct combinations must have:

$P$ as a methyl styrene derivative giving tolualdehyde on ozonolysis.
$R$ as $\alpha$ -methyl styrene giving acetophenone on ozonolysis.

These correspond to options A and B.

Comparison with stored correct answer

Stored correct answer: A, B

Derived answer: A, B

They match.

More from Aldehydes Ketones and Carboxylic Acids

View written solutionFree

Correct answer: A, B

Identify the nature of $Q$ and $S$ from the given tests

Both $Q$ and $S$ have molecular formula $C_8H_8O$ .

For degree of unsaturation:

\text{DU} = \frac{2C+2-H}{2} = \frac{2(8)+2-8}{2} = \frac{10}{2} = 5

So each compound has total unsaturation $=5$ , which strongly suggests an aromatic ring $(4)$ plus one more unsaturation, most likely a carbonyl.

Thus $Q$ and $S$ are most likely aromatic carbonyl compounds with formula $C_8H_8O$ .

Use Cannizzaro and haloform tests

For $Q$ :

$Q$ undergoes Cannizzaro reaction $\Rightarrow$ it must be an aldehyde without $\alpha$ -hydrogen.
$Q$ does not undergo haloform reaction $\Rightarrow$ it is not a methyl ketone and not acetaldehyde-type.

Among aromatic carbonyl compounds with formula $C_8H_8O$ , the suitable aldehyde is:

C_6H_5CH_2CHO \quad \text{(phenylacetaldehyde)}

but this has $\alpha$ -hydrogen, so it does not give Cannizzaro.

Another possibility is:

CH_3C_6H_4CHO \quad \text{(tolualdehyde)}

This is an aromatic aldehyde where the carbonyl carbon is directly attached to the ring, so there is no $\alpha$ -hydrogen on the carbon adjacent to the carbonyl. Hence it gives Cannizzaro reaction.

Also it is not a methyl ketone, so it does not give haloform reaction.

Therefore,

Q = CH_3C_6H_4CHO \quad \text{(methyl benzaldehyde / tolualdehyde)}

For $S$ :

$S$ undergoes haloform reaction $\Rightarrow$ it must contain the group

CH_3CO-

$S$ does not undergo Cannizzaro reaction $\Rightarrow$ it is not a non-enolizable aldehyde; a ketone is fine.

With formula $C_8H_8O$ , the aromatic methyl ketone is:

C_6H_5COCH_3 \quad \text{(acetophenone)}

This gives haloform reaction and does not give Cannizzaro.

Therefore,

S = C_6H_5COCH_3 \quad \text{(acetophenone)}

Now infer alkenes $P$ and $R$ whose ozonolysis give $Q$ and $S$ respectively

Ozonolysis cleaves a $C=C$ bond to form carbonyl compounds.

For $Q = CH_3C_6H_4CHO$ :

To get an aldehyde of type Ar-CHO by ozonolysis, the alkene should be of styrene type:

CH_3C_6H_4CH=CH_2

On ozonolysis:

CH_3C_6H_4CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3C_6H_4CHO + HCHO

So a methyl styrene (tolyl ethene) can be $P$ .

For $S = C_6H_5COCH_3$ :

To get acetophenone by ozonolysis, the alkene should be:

C_6H_5C(CH_3)=CH_2

On ozonolysis:

C_6H_5C(CH_3)=CH_2 \xrightarrow[Zn/H_2O]{O_3} C_6H_5COCH_3 + HCHO

So $\alpha$ -methyl styrene can be $R$ .

Conclusion

Thus the correct combinations must have:

$P$ as a methyl styrene derivative giving tolualdehyde on ozonolysis.
$R$ as $\alpha$ -methyl styrene giving acetophenone on ozonolysis.

These correspond to options A and B.

Comparison with stored correct answer

Stored correct answer: A, B

Derived answer: A, B

They match.

Aldehydes Ketones and Carboxylic Acids question

Aldehydes Ketones and Carboxylic Acids question

For $Q$ :

For $S$ :

For $Q = CH_3C_6H_4CHO$ :

For $S = C_6H_5COCH_3$ :

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Aldehydes Ketones and Carboxylic Acids question

Aldehydes Ketones and Carboxylic Acids question

For $Q$ :

For $S$ :

For $Q = CH_3C_6H_4CHO$ :

For $S = C_6H_5COCH_3$ :

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For QQQ:

For SSS:

For Q=CH3C6H4CHOQ = CH_3C_6H_4CHOQ=CH3​C6​H4​CHO:

For S=C6H5COCH3S = C_6H_5COCH_3S=C6​H5​COCH3​:

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For QQQ:

For SSS:

For Q=CH3C6H4CHOQ = CH_3C_6H_4CHOQ=CH3​C6​H4​CHO:

For S=C6H5COCH3S = C_6H_5COCH_3S=C6​H5​COCH3​:

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For $Q$ :

For $S$ :

For $Q = CH_3C_6H_4CHO$ :

For $S = C_6H_5COCH_3$ :

For $Q$ :

For $S$ :

For $Q = CH_3C_6H_4CHO$ :

For $S = C_6H_5COCH_3$ :