The compound if contains an even number 'n' of chiral carbons, but the molecule can be divided into two equal and si,ilar halves , then how many optical active form it will have :

2^{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{(n - 1)}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^{(n - 1) / 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{(n - 1)} - 2^{(n - 1) / 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2^{(n - 1)}$
If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then $2^{n - 1}$ is the number of optically active forms.

Suggest Corrections

Similar questions

Q. Assertion :If 'n' is even then

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

Reason:

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

Q. If

a = \infty \sum n = 1 \frac{2 n}{(2 n - 1)!}, b = \infty \sum n = 1 \frac{2 n}{(2 n + 1)!}

, then

a b

equals

Q. Prove that

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

where n is a + ive integer.

How many of the following statements are true?
(1) $r^{n} C_{r} = n \times^{n - 1} C_{r - 1}$

(2) Each term of the sum
$(^{2 n} C_{1})^{2} + 2 \times (^{2 n} C_{2})^{2} + 3. (^{2 n} C_{3})^{2} . . . . . + 2 n . (^{2 n} C_{2 n})^{2}$ can be written as $r (^{2 n} C_{r})^{2}$
(3) coefficient of $x^{2 n - 1}$ in
$(^{2 n - 1} C_{0} +^{2 n - 1} C_{1} x +^{2 n - 1} C_{2} x^{2} . . . . .^{2 n - 1} C^{2 n - 1} x^{2 n - 1}) \times (^{2 n} C_{0} +^{2 n} C_{1} x +^{2 n} C_{2} x^{2} . . . . .^{2 n} C_{2 n} x^{2 n})$
= $^{2 n - 1} C_{0} \times^{2 n} C_{2 n - 1} +^{2 n - 1} C_{1} \times^{2 n} C_{2 n - 2} . . . . .^{2 n - 1} C_{2 n - 1} .^{2 n} C_{0}$

___

Q. If

(1 + x + x^{2})^{n} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}

, then

2 a_{1} - 3 a_{2} + . . . - (2 n + 1) a_{2 n}

is equal to

Join BYJU'S Learning Program

The compound if contains an even number 'n' of chiral carbons, but the molecule can be divided into two equal and si,ilar halves , then how many optical active form it will have :

The correct option is B 2(n−1)If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then 2n−1 is the number of optically active forms.

The correct option is B $2^{(n - 1)}$
If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then $2^{n - 1}$ is the number of optically active forms.