How many asymmetric carbon atoms are present in the following compound? CH3 CH2 CHCl CHCl CH2 CH2OH

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We start by restating the question: How many asymmetric (chiral) carbon atoms are present in the compound CH3 CH2 CHCl CHCl CH2 CH2OH?

First, examine each carbon atom in the chain to identify potential stereocenters. A carbon is stereogenic if it is bonded to four different substituents.

- Carbon-3 (the first CHCl group): This carbon is bonded to four groups: a hydrogen, a chlorine, the preceding carbon fragment (CH2–CH3), and the following carbon fragment (the second CHCl–CH2–CH2OH). The two carbon substituents are chemically distinct (one side leads toward a CH3 end, the other toward a halogenated center), so Carbon-3 is a stereocenter.

- Carbon-4 (the second CHCl group): This carbon is bonded to four groups: a hydrogen, a chlorine, the preceding fragment (the first CHCl–CH2–CH3 region), and the following fragment (CH2–CH2OH). The two carbon substituents are different, so Carbon-4 is also a stereocenter.

- Carbons 1, 2, 5, and 6 (the terminal and middle carbons): Carbon-1 (CH3) and Carbon-2 (CH2) lack four distinct substituents (they are not attached to four different groups each). Carbon-5 (CH2) and Carbon-6 (CH2OH) are similarly not tetra-substituted into four unique groups. Therefore, they cannot be stereocenters.

Thus, there are two stereogenic centers in this molecule, located at the two CHCl-bearing carbons in the chain. The overall symmetry of the molecule does not create a meso form here because the terminal groups differ (CH3 at one end and CH2OH at the other), so there isn’t an internal plane of symmetry that could render a meso compound.

In summary, the compound contains two asymmetric carbon atoms, corresponding to two stereocenters in the molecule.

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How many asymmetric carbon atoms are present in the following compound? CH3 CH2 CHCl CHCl CH2 CH2OH

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