HC - CH 2CH3MgBr-A B C - DIf the Molecular weight of D is x gm- then write your answer as x2

HC ≡ CH BrMg C ≡ C MgBr HOOC C ≡ C COOH H_ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Addition of Alcohols and Carboxylic Acids for Alkynes

HC ≡ CH 2CH3M...

Question

$H C \equiv C H 2 C H_{3} M g B r ⟶ (A) (2) 2 C O_{2} - ------- \to (i i) H O H / H^{\oplus} (B) H_{2} S O_{4} - --- \to H g S O_{4} (C) Δ \to (D)$
If the Molecular weight of D is $x g m$ , then write your answer as $\frac{x}{2}$

Open in App

Solution

$H C \equiv C H 2 C H_{3} M g B r - ------ \to - 2 C H_{4} B r M g - C \equiv C - M g B r (i) 2 C O_{2} - ------ \to (i i) H_{2} O / H^{\oplus} H O O C - C \equiv C - C O O H H_{2} S O_{4} ↓ H g S O_{4} H O O C - O | | C - C H_{3} - C O_{2} \leftarrow -- - Δ H O O C - β C | | O - α C H_{2} - C O O H$

Molecular weight of the product (x) = 88
$∴ \frac{x}{2} = 44$

Suggest Corrections

Similar questions

H C \equiv C H 2 C H_{3} M g B r ⟶ (A) (2) 2 C O_{2} - ------- \to (i i) H O H / H^{\oplus} (B) H_{2} S O_{4} - --- \to H g S O_{4} (C) Δ \to (D)

If the Molecular weight of D is

x g m

, then write your answer as

\frac{x}{2}

H C \equiv C H 2 C H_{3} M g B r ⟶ (A) (2) 2 C O_{2} - ------- \to (i i) H O H / H^{\oplus} (B) H_{2} S O_{4} - --- \to H g S O_{4} (C) Δ \to (D)

If the Molecular weight of D is

x g m

, then write your answer as

\frac{x}{2}

Q. If weight of

1

mango is

250

gms, then calculate the weight of

16

mangoes. (Write your answer in kilograms)

At STP weight of $1 l i t r e$ of a gas is $3.16 g r a m s$ and weight of $1 l i t r e$ of Hydrogen is $0.09 g m$ . If molecular weight of Hydrogen is 2, molecular weight of the gas is

Q. Write

\sqrt{7} y = 2 x

as a linear equation of the form ax + by + c = 0. Also, write the values corresponding to a, b, c. Does the graph of this linear equation pass through origin? Give your answer in 'yes' or 'no'.

Join BYJU'S Learning Program

HC≡CH2CH3MgBr⟶(A) (2)2CO2−−−−−−−−→(ii)HOH/H⊕ (B) H2SO4−−−−→HgSO4 (C) Δ→ (D) If the Molecular weight of D is x gm, then write your answer as x2

HC≡CH2CH3MgBr−−−−−−−→−2CH4BrMg−C≡C−MgBr(i)2CO2−−−−−−−→(ii)H2O/H⊕HOOC−C≡C−COOH H2SO4↓HgSO4 HOOC−O||C−CH3−CO2←−−−ΔHOOC−βC||O−αCH2−COOH Molecular weight of the product (x) = 88 ∴ x2=44

$H C \equiv C H 2 C H_{3} M g B r ⟶ (A) (2) 2 C O_{2} - ------- \to (i i) H O H / H^{\oplus} (B) H_{2} S O_{4} - --- \to H g S O_{4} (C) Δ \to (D)$
If the Molecular weight of D is $x g m$ , then write your answer as $\frac{x}{2}$