In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge? 

Correct answer: 1)

Concept:

• Fluorine is the most electronegative element.
• As we move right in a period, the electronegativity keeps increasing up to the halogens.
• Electronegativity is a property of bonded atom that describes its ability to attract shared electrons pair.
• More electronegative atoms have a stronger pull on electrons and release more energy when they gain electrons.
• A carbocation is short of electrons and has a positive charge, where the carbon itself is capable of accepting another two.
• This makes it a lewis acid and makes a carbocation different from other cations.

Explanation:

• The order of electronegativity of attached groups are as follows: Cl>Br>C>Mg.
• Hence, *CH3—CH2 —Cl has the greatest positive charge. As Cl is the most electronegative element among Cl, Mg, and C thus, it has greater -I effect and the asterisk marked carbon in *CH3—CH2 —Cl is expected to have greater positive charge.
• When a more electronegative atom is attached to carbon it causes polarisation of σ−bond due to which, partial positive charge is developed on the carbon atom.
• This polarisation is passed to adjacent carbon atoms with decreased intensity.
• So, more the electronegative atom attached, more will be the positive charge developed on neighbouring carbon.
• Polarisation in presence of metal (option 2), takes place in such a way that, it creates partial negative charge on neighbouring labelled carbon.
• There polarisation does not take place because no electronegative atom (option 4) is attached to carbon.
• So, no partial charge development occurs here on labelled carbon.
• A carbon marked with an asterisk in compound 1 will have the most positive charge among all options because chlorine atoms are of higher electronegativity than bromine atoms.

Conclusion:

Thus, *CH3—CH2 —Cl is expected to have greatest positive charge.