MHT CET · Chemistry · Aldehydes and Ketones
What is the product obtained in the reaction?
\(\text{CH} _3- \text{CH} = \text{CH} - \text{CH} _2- \text{CHO} \xrightarrow{\substack{\text { (i) } \text{LiAlH} _4 \\ \text { (ii) } \text{H} _3 \text{O} ^{+}}}\) product
- A \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}\)
- B \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{COOH}\)
- C \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CHO}\)
- D \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3\)
Answer & Solution
Correct Answer
(A) \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}\)
Step-by-step Solution
Detailed explanation
The advantage of \(\mathrm{LiAlH}_4\) is that it does not reduce the isolated olefinic bond and hence, it can reduce unsaturated aldehydes and ketones to unsaturated alcohols.
\(\text{CH} _3- \text{CH} = \text{CH} - \text{CH} _2- \text{CHO} \xrightarrow{\substack{\text { (i) } \text{LiAlH} _4 \\ \text { (ii) } \text{H} _3 \text{O} ^{+}}}\) \(\text{CH} _3-\text{ CH} = \text{CH} - \text{CH} _2- \text{CH} _2- \text{OH}\)
\(\text{CH} _3- \text{CH} = \text{CH} - \text{CH} _2- \text{CHO} \xrightarrow{\substack{\text { (i) } \text{LiAlH} _4 \\ \text { (ii) } \text{H} _3 \text{O} ^{+}}}\) \(\text{CH} _3-\text{ CH} = \text{CH} - \text{CH} _2- \text{CH} _2- \text{OH}\)
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