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luda_lava [24]
3 years ago
15

Tertiary alkyl halides show SN1 reactions mostly...... Why?

Chemistry
1 answer:
ankoles [38]3 years ago
6 0

Answer:

Generally SN1 reactions are carried out in polar protic solvents. The energy required for breaking the C-X bond is obtained through solvation of halide ion with the proton of the solvent. Tertiary alkyl halides undergo SN1 reaction very fast because of the high stability of tertiary carbocations.

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5 0
2 years ago
A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of OH^-ions is greater. The remaining concentration of [OH^-]ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

[HNO_3] = 0.200 M

[Ca(OH)_2] =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of H^+ ion from both the acids

moles of H^+ in HCl

HCl\rightarrow {H^+}+Cl^-

if 1 L of HClsolution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 \times0.1 moles= 0.005 moles    (1L=1000mL)

moles of H^+ in HCl = 0.005 moles

Similarliy

moles of H^+ in HNO_3

HNO_3\rightarrow H^++NO_3^-}

If 1L of HNO_3 solution= 0.200 moles

Then 0.1L of HNO_3 solution= 0.1 \times 0.200 moles= 0.02 moles

moles of H^+ in HNO_3 =0.02 moles

so, Total moles of H^+ ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of [OH^-] ion from both the bases

Moles of OH^-\text{ in }Ca(OH)_2

Ca(OH)_2\rightarrow Ca^2{+}+2OH^-

1 L of Ca(OH)_2= 0.0100 moles

Then in 0.5 L Ca(OH)_2 solution = 0.5 \times0.0100 moles = 0.005 moles

Ca(OH)_2 produces two moles of OH^- ions

moles of OH^- = 0.005 \times 2= 0.01 moles

Moles of OH^- in RbOH

RbOH\rightarrow Rb^++OH^-

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 \times 0.100 moles = 0.02 moles

Moles of OH^- = 0.02 moles

so,Total moles of OH^- ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both H^+\text{ and }OH^- ions

One mole of H^+ ions will combine with one mole of OH^- ions, so

Total moles of H^+ ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of OH^- ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of H^+ ions = total moles of OH^- ions

0.025 moles H^+ will neutralize the 0.025 moles of OH^-

Moles of OH^- ions is in excess        (from 1 and 2)

The remaining moles of OH^- will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of OH^- ions = \frac{\text{Moles remaining}}{\text{Total volume}}

[OH^-]=\frac{0.005}{0.85}=0.0058M

6 0
4 years ago
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