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The number of optically active optical isomers of the compound is:
The number of optically active isomers = 2n−1−2n−1/2 =23−1−23−1/2 =2.
Find the number of stereoisomers for CH3 – CHOH – CH = CH – CH3.
The number of stereoisomers for CH3 – CHOH – CH = CH – CH3 is four. This is calculated by the formula 2n+1. where n is number of chiral centres, which is 1 in this case. So, 21+1 = 22= 4
A hydrocarbon X is optically. X upon hydrogenation gives an optically inactive alkane Y. Which of the following pair of compounds can be X and Y respectively?
The correct answer is Option B.
The optically active hydrocarbon X is 3-methyl-1-pentene CH2=CH−CH(CH3)CH2CH3. On catalytic hydrogenation, it forms 3-methyl pentane CH3CH2CH(CH3)CH2CH3, which is optically inactive.
What is wrong about enantiomers of 2-chloropropanoic acid?
The correct answer is option D.
Pair of enantiomers react differently with pure enantiomers of other compounds.
How many different stereoisomers exist for the compound below ?
If you name the C attached to Me group as C1 and naming the rest clockwise, you observe that at C1 and C2 there are chiral centers and C5 and C2 are symmetrical. So the two chiral centers produce 2×2=4 optically active(and distinguishable) compounds. Hence, B is the answer.
The correct answer is Option B.
5 isomers are possible. C5H10(CnH2n). Molecules having the CnH2n formulas are most likely to be cyclic alkanes. The five isomers are:
(1) Cyclopentane
(2) 1-Methylcyclobutane
(3) 1-Ethylcyclopropane
(4) 1,1-Dimethylcyclopropane
(5) 1,2-Dimethylcyclopropane
How the following pair of isomers cannot be distinguished ?
Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4
(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are
The general formula of isomeric ketone having molecular mass 100 is C6H12O [6×12+12×1+16].
The possible structure will be :
The number of ketones that gives racemic mixture with NaBH4 is 5 as the ketone with chiral center will give diastereoisomer with NaBH4
Direction (Q. Nos. 10-14) This section contains 5 multiple choice questions. Each question has four
choices (a), (b), (c), and (d), out of which ONE or MORE THAN ONE is correct.
Q.Consider the following molecule
The correct statement concerning the above molecule is/are
It's meso form upon ozonolysis followed by Zn-hydrolysis gives racemic mixture
It's optically active isomers, each upon ozonolysis followed by Zn-hydrolysis gives a single enantiomer.
Consider the following free radical chlorination reaction.
The correct statement about product(s) is/are
There are 4 final products, they are as follow:-
2 products of the structure above due to presence of a chiral carbon.
1 product of structure above due to absence of Chiral Carbon.
1 product of structure above due to absence of Chiral Carbon.
Hence, Options A, B are correct.
The correct statement(s) regarding Isomers of a compound with general name trichlorocyclobutane is/are
The correct statement (s) about the compound
is/are
[IIT JEE 2009]
The correct options are A & D
Since a carbon-carbon double bond is present, the molecule can have either cis isomer or trans isomer.
In cis isomer (which is symmetrical), the number of enantiomers is
2(n−1)=2(2−1)=2
The number of meso compounds possible is 2(n/2)−1=2(2/2)−1=1
Hence, the total number of optical isomers possible is 2+1=3.
Trans isomer is also symmetrical.
Hence, the total number of optical isomers possible for trans isomers is also 3.
Hence, the total number of stereoisomers possible for X is 3+3=6.
Hence, option A is correct and option B is incorrect.
If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4.
Thus, the option C is incorrect.
If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2.
Thus the option D is correct.
The correct statement(s) about the compound given below is/are
The correct answer is option A & D
The given image is Self-Explanatory.
We know that following are the cases when a compound is Optically Inactive:
1. When it does not have a chiral Carbon Atom eg.: Methyl bromide
2. When it has a Plane or Axis of Symmetry eg.: Tartaric Acid (these compounds are called as Meso Compounds)
3. When it forms a Racemic Mixture eg.: Equimolar mixture of d-Lactic Acid and l-Lactic Acid.
The given compound in the question possesses an Axis of Symmetry (shown in the attached image). Hence, it is Optically Inactive.
Which of the given statement (s) about N, O, P, and Q with respect to M is (are) correct?
The Fischer Projections of the given compounds are below.
The number of optical isomers possible for 2, 3-pentanediol is:
2 optical centers.
Total optically active isomers =22
=4
The minimum number of C atoms required for a hydrocarbon to exhibit optical isomerism:
The minimum number of C atoms required for a hydrocarbon to exhibit optical isomerism = 7
Which of the following statements is/are correct regarding the below compound?
Correct option is
A: It is optically inactive due to the plane of symmetry &
B: It is optically inactive due to the center of symmetry
The presence of any kind of symmetry makes the compound optically inactive.
Which of the following are not functional isomers of each other?
Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar or different functional groups.
10 amine to different 10 amines are not functional isomers.
Hence, option ′d′ is the answer.
A pair of enantiomers is possible for _______ isomer of 2,2-dibromobicyclo [2.2.1] heptane.
Only one pair of enantiomers is possible for cis-2,2-dibromobicyclo [2.2.1] heptane. The trans arrangement of one carbon bridge is structurally impossible. Such a molecule would have too much strain.
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