If a compound has "n" asymmetric carbon atoms with different terminal groups, the number of stereoisomers is given by the formula :

(\frac{1}{2})^{n}

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2^{n}

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2 \sqrt{n}

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\sqrt{2}

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Solution

The correct option is A $2^{n}$
The number of stereoisomers is given by the formula $2^{n}$ where "n" is the total number of chiral centers the molecule.

Option B is correct.

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Q. Prove that

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

where n is a + ive integer.

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If a compound has "n" asymmetric carbon atoms with different terminal groups, the number of stereoisomers is given by the formula :

The correct option is A 2nThe number of stereoisomers is given by the formula 2n where "n" is the total number of chiral centers the molecule.Option B is correct.

The correct option is A $2^{n}$
The number of stereoisomers is given by the formula $2^{n}$ where "n" is the total number of chiral centers the molecule.

Option B is correct.