Answer: A) Nonmetal carbon shares valence electrons with each nonmetal chlorine forming four covalent bonds.
Explanation: Covalent bond is formed by sharing of electrons between atoms.
Ionic bond is formed by transfer of electrons between atoms.
Carbon with atomic no 6 and has configuration of . Carbon has 4 valence electrons. It can only share electrons as it is difficult to gain or lose 4 electrons to complete it's octet.
Chlorine with atomic no 17 has configuration of . It has 7 valence electrons and need one electron to complete its octet.
Thus carbon will share 4 electrons, one each with four chlorine atoms to form carbon tetra chloride.
Answer: 0.028 grams
Explanation:
Depression in freezing point :
Formula used for lowering in freezing point is,
or,
where,
= change in freezing point
= freezing point constant (for benzene} =
m = molality
Putting in the values we get:
0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.
Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
