Question

① H2O ② Red hot fetube ③ CH3COCl/AlCl3 $\longrightarrow$ P ④ NaNO2 ⑤ NH3,$\triangle$ PhCONH2 (xgm) ⑥ NaOBr PhNH2 ⑦ HNO2 (70%) PhN2 ⑧ CuCN/KCN ⑨ CH3MgBr then H2O ⑩ Br2/NaOH (80%) ⑪ NaOH, CaO/$\triangle$

Answer 1

121

Answer 2

The question asks to identify the product 'P' from a given reaction and then refers to "PhCONH2 (xgm)". Given the ambiguous phrasing and lack of initial quantities for a calculation, "xgm" is most likely asking for the molecular weight of PhCONH2.

Let's break down the problem:

Part 1: Identify P

The reaction sequence states: "③ CH3COCl/AlCl3 $\longrightarrow$ P"

This is a Friedel-Crafts acylation reaction. Assuming the starting material is benzene (PhH), which is a common aromatic reactant when not specified:

Benzene reacts with acetyl chloride (CH3COCl) in the presence of anhydrous aluminium chloride (AlCl3) to form acetophenone.

The reaction is:

$C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$

Therefore, P is Acetophenone ($C_6H_5COCH_3$).

Part 2: Determine 'x' in PhCONH2 (xgm)

"PhCONH2 (xgm)" refers to benzamide. Since no initial quantities are provided to calculate a mass, 'x' most likely represents the molecular weight of benzamide.

The molecular formula of benzamide is $C_7H_7NO$.

Using atomic masses: C = 12 u, H = 1 u, N = 14 u, O = 16 u.

Molar mass of $C_7H_7NO = (7 \times 12) + (7 \times 1) + (1 \times 14) + (1 \times 16)$

$= 84 + 7 + 14 + 16$

$= 121 \text{ g/mol}$

So, if 'x' represents the molecular weight, then x = 121.

Overall Reaction Sequence (to produce PhCONH2 using given reagents):

While not explicitly asked as a single sequence, understanding how PhCONH2 can be formed from the given reagents helps to contextualize the problem.

1. Formation of Acetophenone (P):

   Benzene $\xrightarrow{③ CH_3COCl/AlCl_3}$ $C_6H_5COCH_3$ (Acetophenone, P)

2. Oxidation of Acetophenone to Sodium Benzoate (Haloform reaction):

   $C_6H_5COCH_3 \xrightarrow{⑩ Br_2/NaOH \ (80\%)}$ $C_6H_5COONa + CHBr_3$

3. Acidification to Benzoic Acid:

   $C_6H_5COONa \xrightarrow{① H_2O \ (acidic \ workup)}$ $C_6H_5COOH$ (Benzoic acid)

4. Formation of Benzamide:

   $C_6H_5COOH \xrightarrow{⑤ NH_3, \triangle}$ $C_6H_5CONH_2$ (Benzamide)

The question is a mix of identifying an intermediate product (P) and a value related to a final product (PhCONH2).