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ALDEHYDES, KETONES AND CARBOXYLIC ACID

26608 An alkene of molecular formula $C_{9} H_{18}$ on ozonolysis gives $2, 2$ dimethyl propanal and $2 -$ butanon, then the alkene is

1

2, 2, 4 -

trimethyl

- 3 -

hexene

2

2, 2, 6 -

trimethyl

- 3 -

hexene

3

2, 3, 4 -

trimethyl

- 2 -

hexene

4

2, 2, 4 -

trimethyl

- 2 -

hexene

Explanation:

$C_{9} H_{18} + O_{3} \to$ $\underset{2, 2 dimethyl propanal}{\begin{matrix} \begin{matrix} C H_{3} \\ | \end{matrix} \\ H_{3} C - C - C H O \\ | \\ C H_{3} \end{matrix}}$ $+ \underset{2 - butanone}{C H_{3} C H_{2} C O C H_{3}}$
On the basis of product formation, it would be alkene
$\underset{2, 2, 4 - trimethyl - 3 - hexene}{\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3}}$
$\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3} \overset{O_{3}}{\to}$ 357-s43

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26609 When m-chlorobenzaldehyde is treated with $50 %$ $K O H$ solution, the product $(s)$ obtained is (are)

1

2

3

4

Explanation:

(c) It is cannizzaro reaction $- 2$ 358-s2

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26610 $A$ and $B$ in the following reactions are

1

2

3

4

A = R R^{'} C H_{2} C N, B = N a O H

Explanation:

(a) $R - C O - R^{'} \overset{H C N}{\to}$ - $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - R^{'} \\ | \\ C N \end{matrix} \overset{L i A l H_{4}}{\to}$ $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - C H_{2} N H_{2} \\ | \\ R^{'} \end{matrix}$

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26611 Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

1

Z n

amalgam and

H C l

acid

2

P d / B a S O_{4}

3 Anhydrous

A l C l_{3}

4

N i / P t

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26608 An alkene of molecular formula $C_{9} H_{18}$ on ozonolysis gives $2, 2$ dimethyl propanal and $2 -$ butanon, then the alkene is

1

2, 2, 4 -

trimethyl

- 3 -

hexene

2

2, 2, 6 -

trimethyl

- 3 -

hexene

3

2, 3, 4 -

trimethyl

- 2 -

hexene

4

2, 2, 4 -

trimethyl

- 2 -

hexene

Explanation:

$C_{9} H_{18} + O_{3} \to$ $\underset{2, 2 dimethyl propanal}{\begin{matrix} \begin{matrix} C H_{3} \\ | \end{matrix} \\ H_{3} C - C - C H O \\ | \\ C H_{3} \end{matrix}}$ $+ \underset{2 - butanone}{C H_{3} C H_{2} C O C H_{3}}$
On the basis of product formation, it would be alkene
$\underset{2, 2, 4 - trimethyl - 3 - hexene}{\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3}}$
$\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3} \overset{O_{3}}{\to}$ 357-s43

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26609 When m-chlorobenzaldehyde is treated with $50 %$ $K O H$ solution, the product $(s)$ obtained is (are)

1

2

3

4

Explanation:

(c) It is cannizzaro reaction $- 2$ 358-s2

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26610 $A$ and $B$ in the following reactions are

1

2

3

4

A = R R^{'} C H_{2} C N, B = N a O H

Explanation:

(a) $R - C O - R^{'} \overset{H C N}{\to}$ - $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - R^{'} \\ | \\ C N \end{matrix} \overset{L i A l H_{4}}{\to}$ $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - C H_{2} N H_{2} \\ | \\ R^{'} \end{matrix}$

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26611 Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

1

Z n

amalgam and

H C l

acid

2

P d / B a S O_{4}

3 Anhydrous

A l C l_{3}

4

N i / P t

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26608 An alkene of molecular formula $C_{9} H_{18}$ on ozonolysis gives $2, 2$ dimethyl propanal and $2 -$ butanon, then the alkene is

1

2, 2, 4 -

trimethyl

- 3 -

hexene

2

2, 2, 6 -

trimethyl

- 3 -

hexene

3

2, 3, 4 -

trimethyl

- 2 -

hexene

4

2, 2, 4 -

trimethyl

- 2 -

hexene

Explanation:

$C_{9} H_{18} + O_{3} \to$ $\underset{2, 2 dimethyl propanal}{\begin{matrix} \begin{matrix} C H_{3} \\ | \end{matrix} \\ H_{3} C - C - C H O \\ | \\ C H_{3} \end{matrix}}$ $+ \underset{2 - butanone}{C H_{3} C H_{2} C O C H_{3}}$
On the basis of product formation, it would be alkene
$\underset{2, 2, 4 - trimethyl - 3 - hexene}{\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3}}$
$\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3} \overset{O_{3}}{\to}$ 357-s43

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26609 When m-chlorobenzaldehyde is treated with $50 %$ $K O H$ solution, the product $(s)$ obtained is (are)

1

2

3

4

Explanation:

(c) It is cannizzaro reaction $- 2$ 358-s2

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26610 $A$ and $B$ in the following reactions are

1

2

3

4

A = R R^{'} C H_{2} C N, B = N a O H

Explanation:

(a) $R - C O - R^{'} \overset{H C N}{\to}$ - $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - R^{'} \\ | \\ C N \end{matrix} \overset{L i A l H_{4}}{\to}$ $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - C H_{2} N H_{2} \\ | \\ R^{'} \end{matrix}$

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26611 Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

1

Z n

amalgam and

H C l

acid

2

P d / B a S O_{4}

3 Anhydrous

A l C l_{3}

4

N i / P t

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26608 An alkene of molecular formula $C_{9} H_{18}$ on ozonolysis gives $2, 2$ dimethyl propanal and $2 -$ butanon, then the alkene is

1

2, 2, 4 -

trimethyl

- 3 -

hexene

2

2, 2, 6 -

trimethyl

- 3 -

hexene

3

2, 3, 4 -

trimethyl

- 2 -

hexene

4

2, 2, 4 -

trimethyl

- 2 -

hexene

Explanation:

$C_{9} H_{18} + O_{3} \to$ $\underset{2, 2 dimethyl propanal}{\begin{matrix} \begin{matrix} C H_{3} \\ | \end{matrix} \\ H_{3} C - C - C H O \\ | \\ C H_{3} \end{matrix}}$ $+ \underset{2 - butanone}{C H_{3} C H_{2} C O C H_{3}}$
On the basis of product formation, it would be alkene
$\underset{2, 2, 4 - trimethyl - 3 - hexene}{\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3}}$
$\overset{C H_{3}}{\overset{|}{\underset{C H_{3}}{\underset{|}{C H_{3} - C - H C}}}} = \overset{C H_{3}}{\overset{|}{C -}} C H_{2} C H_{3} \overset{O_{3}}{\to}$ 357-s43

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26609 When m-chlorobenzaldehyde is treated with $50 %$ $K O H$ solution, the product $(s)$ obtained is (are)

1

2

3

4

Explanation:

(c) It is cannizzaro reaction $- 2$ 358-s2

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26610 $A$ and $B$ in the following reactions are

1

2

3

4

A = R R^{'} C H_{2} C N, B = N a O H

Explanation:

(a) $R - C O - R^{'} \overset{H C N}{\to}$ - $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - R^{'} \\ | \\ C N \end{matrix} \overset{L i A l H_{4}}{\to}$ $\begin{matrix} \begin{matrix} O H \\ | \end{matrix} \\ R - C - C H_{2} N H_{2} \\ | \\ R^{'} \end{matrix}$

ALDEHYDES, KETONES AND CARBOXYLIC ACID

26611 Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

1

Z n

amalgam and

H C l

acid

2

P d / B a S O_{4}

3 Anhydrous

A l C l_{3}

4

N i / P t