28234
In the given reaction :$\begin{array}{*{20}{c}} O \\ {||} \\ {{C_6}{H_5} - C - C{H_3}} \end{array}\mathop {\xrightarrow{{(i)\,B{r_2}/KOH}}}\limits_{(ii)\,{H^ \oplus }} CHB{r_3} + [X]$\([X]\) will be :
1 \(C_6H_5-CHO\)
2 \(C_6H_5COOH\)
3 \(C_6H_5-CH_2OH\)
4 \(CH_3COOH\)
Explanation:
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28235
In the reaction sequence :\(CH_3 -C\equiv C-H\) \(\xrightarrow{{C{H_3}MgBr}}\) \(CH_4+(A)\) \(\mathop {\xrightarrow{{(i)\,C{O_2}}}}\limits_{(ii)\,{H_2}O/{H^ \oplus }} \) \((B)\)\((B)\) will be :
1 \(CH_3-C\equiv C-CH_3\)
2 \(CH_3-C\equiv C-MgBr\)
3 \(CH_3-C\equiv C-COOH\)
4 \(CH_3-CH=CH-COOH\)
Explanation:
\(I-\)propyne reacts with methyl magnesium bromide to form prop-1ynylmagnesium bromide which is compound \(A\). Methane is the byproduct.The prop\(-1-\)ynyl anion acts as a nucleophile and attacks carbon dioxide. This is followed by hydrolysis to form but\(-2-\)ynoic acid which is compound \(B\). \(CH _3- C \equiv C - H \stackrel{ CH _3 MgBr }{\longrightarrow} CH _4+ CH _3- C \equiv C - MgBr \underset{\text { (ii) } H _2 O / H ^{\oplus}}{\stackrel{( i ) CO _2}{\longrightarrow}} CH _3- C \equiv \underset{\text { (B) }}{ C }- COOH\)
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28236
Consider the given reaction\(RCOOAg \xrightarrow{{B{r_2}/\Delta }} R-Br\)which one of the following acid will give maximum yield of \(R-Br\) in the above reaction?
\(R C O O A g \stackrel{ Br _2 / \triangle}{\longrightarrow} R-B r\) This reaction is a name reaction known as Hunsdiccker reaction. It follows free radical mechanism. Reactivity rate follows the order \(1^{\circ} > 2^{\circ} > 3^{\circ}\).
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28237
Benzoic acid on treatment with hydrazoic acid in the presence of concentrated sulphuric acid gives:
1 Benzamide
2 Sodium benzoate
3 Aniline
4 \(C_6H_5CON_3\)
Explanation:
Benzoic acid on treatment with hydrazoic acid (N \(_{3} \mathrm{H}\) ) in the presence of concentrated sulphuric acid gives aniline. This reaction is called Schmidt reaction. It involves an isocyanate intermediate. Option C is correct.
28234
In the given reaction :$\begin{array}{*{20}{c}} O \\ {||} \\ {{C_6}{H_5} - C - C{H_3}} \end{array}\mathop {\xrightarrow{{(i)\,B{r_2}/KOH}}}\limits_{(ii)\,{H^ \oplus }} CHB{r_3} + [X]$\([X]\) will be :
1 \(C_6H_5-CHO\)
2 \(C_6H_5COOH\)
3 \(C_6H_5-CH_2OH\)
4 \(CH_3COOH\)
Explanation:
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28235
In the reaction sequence :\(CH_3 -C\equiv C-H\) \(\xrightarrow{{C{H_3}MgBr}}\) \(CH_4+(A)\) \(\mathop {\xrightarrow{{(i)\,C{O_2}}}}\limits_{(ii)\,{H_2}O/{H^ \oplus }} \) \((B)\)\((B)\) will be :
1 \(CH_3-C\equiv C-CH_3\)
2 \(CH_3-C\equiv C-MgBr\)
3 \(CH_3-C\equiv C-COOH\)
4 \(CH_3-CH=CH-COOH\)
Explanation:
\(I-\)propyne reacts with methyl magnesium bromide to form prop-1ynylmagnesium bromide which is compound \(A\). Methane is the byproduct.The prop\(-1-\)ynyl anion acts as a nucleophile and attacks carbon dioxide. This is followed by hydrolysis to form but\(-2-\)ynoic acid which is compound \(B\). \(CH _3- C \equiv C - H \stackrel{ CH _3 MgBr }{\longrightarrow} CH _4+ CH _3- C \equiv C - MgBr \underset{\text { (ii) } H _2 O / H ^{\oplus}}{\stackrel{( i ) CO _2}{\longrightarrow}} CH _3- C \equiv \underset{\text { (B) }}{ C }- COOH\)
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28236
Consider the given reaction\(RCOOAg \xrightarrow{{B{r_2}/\Delta }} R-Br\)which one of the following acid will give maximum yield of \(R-Br\) in the above reaction?
\(R C O O A g \stackrel{ Br _2 / \triangle}{\longrightarrow} R-B r\) This reaction is a name reaction known as Hunsdiccker reaction. It follows free radical mechanism. Reactivity rate follows the order \(1^{\circ} > 2^{\circ} > 3^{\circ}\).
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28237
Benzoic acid on treatment with hydrazoic acid in the presence of concentrated sulphuric acid gives:
1 Benzamide
2 Sodium benzoate
3 Aniline
4 \(C_6H_5CON_3\)
Explanation:
Benzoic acid on treatment with hydrazoic acid (N \(_{3} \mathrm{H}\) ) in the presence of concentrated sulphuric acid gives aniline. This reaction is called Schmidt reaction. It involves an isocyanate intermediate. Option C is correct.
28234
In the given reaction :$\begin{array}{*{20}{c}} O \\ {||} \\ {{C_6}{H_5} - C - C{H_3}} \end{array}\mathop {\xrightarrow{{(i)\,B{r_2}/KOH}}}\limits_{(ii)\,{H^ \oplus }} CHB{r_3} + [X]$\([X]\) will be :
1 \(C_6H_5-CHO\)
2 \(C_6H_5COOH\)
3 \(C_6H_5-CH_2OH\)
4 \(CH_3COOH\)
Explanation:
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28235
In the reaction sequence :\(CH_3 -C\equiv C-H\) \(\xrightarrow{{C{H_3}MgBr}}\) \(CH_4+(A)\) \(\mathop {\xrightarrow{{(i)\,C{O_2}}}}\limits_{(ii)\,{H_2}O/{H^ \oplus }} \) \((B)\)\((B)\) will be :
1 \(CH_3-C\equiv C-CH_3\)
2 \(CH_3-C\equiv C-MgBr\)
3 \(CH_3-C\equiv C-COOH\)
4 \(CH_3-CH=CH-COOH\)
Explanation:
\(I-\)propyne reacts with methyl magnesium bromide to form prop-1ynylmagnesium bromide which is compound \(A\). Methane is the byproduct.The prop\(-1-\)ynyl anion acts as a nucleophile and attacks carbon dioxide. This is followed by hydrolysis to form but\(-2-\)ynoic acid which is compound \(B\). \(CH _3- C \equiv C - H \stackrel{ CH _3 MgBr }{\longrightarrow} CH _4+ CH _3- C \equiv C - MgBr \underset{\text { (ii) } H _2 O / H ^{\oplus}}{\stackrel{( i ) CO _2}{\longrightarrow}} CH _3- C \equiv \underset{\text { (B) }}{ C }- COOH\)
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28236
Consider the given reaction\(RCOOAg \xrightarrow{{B{r_2}/\Delta }} R-Br\)which one of the following acid will give maximum yield of \(R-Br\) in the above reaction?
\(R C O O A g \stackrel{ Br _2 / \triangle}{\longrightarrow} R-B r\) This reaction is a name reaction known as Hunsdiccker reaction. It follows free radical mechanism. Reactivity rate follows the order \(1^{\circ} > 2^{\circ} > 3^{\circ}\).
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28237
Benzoic acid on treatment with hydrazoic acid in the presence of concentrated sulphuric acid gives:
1 Benzamide
2 Sodium benzoate
3 Aniline
4 \(C_6H_5CON_3\)
Explanation:
Benzoic acid on treatment with hydrazoic acid (N \(_{3} \mathrm{H}\) ) in the presence of concentrated sulphuric acid gives aniline. This reaction is called Schmidt reaction. It involves an isocyanate intermediate. Option C is correct.
28234
In the given reaction :$\begin{array}{*{20}{c}} O \\ {||} \\ {{C_6}{H_5} - C - C{H_3}} \end{array}\mathop {\xrightarrow{{(i)\,B{r_2}/KOH}}}\limits_{(ii)\,{H^ \oplus }} CHB{r_3} + [X]$\([X]\) will be :
1 \(C_6H_5-CHO\)
2 \(C_6H_5COOH\)
3 \(C_6H_5-CH_2OH\)
4 \(CH_3COOH\)
Explanation:
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28235
In the reaction sequence :\(CH_3 -C\equiv C-H\) \(\xrightarrow{{C{H_3}MgBr}}\) \(CH_4+(A)\) \(\mathop {\xrightarrow{{(i)\,C{O_2}}}}\limits_{(ii)\,{H_2}O/{H^ \oplus }} \) \((B)\)\((B)\) will be :
1 \(CH_3-C\equiv C-CH_3\)
2 \(CH_3-C\equiv C-MgBr\)
3 \(CH_3-C\equiv C-COOH\)
4 \(CH_3-CH=CH-COOH\)
Explanation:
\(I-\)propyne reacts with methyl magnesium bromide to form prop-1ynylmagnesium bromide which is compound \(A\). Methane is the byproduct.The prop\(-1-\)ynyl anion acts as a nucleophile and attacks carbon dioxide. This is followed by hydrolysis to form but\(-2-\)ynoic acid which is compound \(B\). \(CH _3- C \equiv C - H \stackrel{ CH _3 MgBr }{\longrightarrow} CH _4+ CH _3- C \equiv C - MgBr \underset{\text { (ii) } H _2 O / H ^{\oplus}}{\stackrel{( i ) CO _2}{\longrightarrow}} CH _3- C \equiv \underset{\text { (B) }}{ C }- COOH\)
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28236
Consider the given reaction\(RCOOAg \xrightarrow{{B{r_2}/\Delta }} R-Br\)which one of the following acid will give maximum yield of \(R-Br\) in the above reaction?
\(R C O O A g \stackrel{ Br _2 / \triangle}{\longrightarrow} R-B r\) This reaction is a name reaction known as Hunsdiccker reaction. It follows free radical mechanism. Reactivity rate follows the order \(1^{\circ} > 2^{\circ} > 3^{\circ}\).
ALDEHYDES, KETONES AND CARBOXYLIC ACID
28237
Benzoic acid on treatment with hydrazoic acid in the presence of concentrated sulphuric acid gives:
1 Benzamide
2 Sodium benzoate
3 Aniline
4 \(C_6H_5CON_3\)
Explanation:
Benzoic acid on treatment with hydrazoic acid (N \(_{3} \mathrm{H}\) ) in the presence of concentrated sulphuric acid gives aniline. This reaction is called Schmidt reaction. It involves an isocyanate intermediate. Option C is correct.
