Which of the following molecules is achiral? A) (2R,3R)-2,3-Dichloropentane B) (2R,3S)-2,3-Dichloropentane C) (2S,3S)-2,3-Dichlorobutane D) (2R,3S)-2,3-Dichlorobutane E) None of these

E) None of these

Explanation:

For this question, we must remember the definition of a chiral carbon. In the chiral carbons, we have four different groups around the carbon.

So, an achiral molecule would be a molecule that does not have chiral carbons. That is, in an achiral molecule in all carbons we have at least one repeating group.

In all the molecules, which we have in the question, we have absolute configuration. That is, we have chiral carbons. In molecule A we have two chiral carbons (R and R), in molecule b we have two chiral carbons (R and S), in molecule c we have two chiral carbons (S and S) and in molecule d we have two chiral carbons (R and S). Therefore in all molecules, we have chiral carbons.

Now, if we analyze molecules a and c, we have a plane of symmetry. If we have a plane of symmetry, even though there are chiral carbons, there will be no optical activity. That is, these molecules do not have the ability to deflect polarized light despite having chiral carbons. Therefore molecules a and c are meso compounds.

But optical activity is not related to chirality.

See figure 1 to further explanations

I hope it helps!

b.final

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol