Carbonyl compounds on reduction with selective reducing agents give alcohols. The structure of alcohol formed depends upon the nature of reducing agents. LiAlH 4, NaBH 4, sodium alcohol, Mg ( Hg ) H 2 O etc can be used. When CH 3- CH = CH - CHO is reduced with NaBH 4, the product formed is

Question

Carbonyl compounds on reduction with selective reducing agents give alcohols. The structure of alcohol formed depends upon the nature of reducing agents. LiAlH 4, NaBH 4, sodium alcohol, Mg ( Hg ) H 2 O etc can be used. When CH 3- CH = CH - CHO is reduced with NaBH 4, the product formed is (A) CH 3- CH = CH - CH 2 OH (B) CH 3- CH 2- CH 2- CHO (C) CH 3- CH 2- CH 2- CH 3 (D) CH 3- CH 2- CH 2- CH 2 OH

Tardigrade Admin · Accepted Answer

NaBH 4 does not reduce C = C or C ≡ C bond ; only > C = O group is reduced to alcoholic group. CH 3- CH = CH - CHO xrightarrowNaBH4CH 3- CH = CH - CH 2 OH

$C H_{3} - C H = C H - C H_{2} O H$

$C H_{3} - C H_{2} - C H_{2} - C H O$

$C H_{3} - C H_{2} - C H_{2} - C H_{3}$

$C H_{3} - C H_{2} - C H_{2} - C H_{2} O H$

$N a B H_{4}$ does not reduce $C = C$ or $C \equiv C$ bond ; only $> C = O$ group is reduced to alcoholic group.
$C H_{3} - C H = C H - C H O NaBH X_{4} C H_{3} - C H = C H - C H_{2} O H$

Q. Carbonyl compounds on reduction with selective reducing agents give alcohols. The structure of alcohol formed depends upon the nature of reducing agents. LiAlH4​, NaBH4​, sodium alcohol, Mg(Hg)H2​O etc can be used. When CH3​−CH=CH−CHO is reduced with NaBH4​, the product formed is

CH3​−CH=CH−CH2​OH

CH3​−CH2​−CH2​−CHO

CH3​−CH2​−CH2​−CH3​

CH3​−CH2​−CH2​−CH2​OH

NaBH4​ does not reduce C=C or C≡C bond ; only >C=O group is reduced to alcoholic group. CH3​−CH=CH−CHONaBHX4​​CH3​−CH=CH−CH2​OH

$C H_{3} - C H = C H - C H_{2} O H$

$C H_{3} - C H_{2} - C H_{2} - C H O$

$C H_{3} - C H_{2} - C H_{2} - C H_{3}$

$C H_{3} - C H_{2} - C H_{2} - C H_{2} O H$

$N a B H_{4}$ does not reduce $C = C$ or $C \equiv C$ bond ; only $> C = O$ group is reduced to alcoholic group.
$C H_{3} - C H = C H - C H O NaBH X_{4} C H_{3} - C H = C H - C H_{2} O H$