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kifflom [539]
3 years ago
8

Using only the 1h nmr spectrum of the crude nitration product, determine if the product is mostly ortho-, meta-, or para- substi

tuted. (note: jortho= ~8 hz, jmeta= ~3 hz and jpara = ~1 hz. also aromatic protons ortho to no2 or c=o groups are generally downfield.)
a.major product(s) = _______________________________
b.describe how you came to your conclusi

Chemistry
1 answer:
tatiyna3 years ago
7 0
<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:

</span>

In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.

On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration

The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.

The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.

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8 0
2 years ago
"11. Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. Abigail place
Amanda [17]

Answer:

44 mL of Na2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

Step 2:

Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:

Molarity of Ba(NO3)2 = 0.5 M

Volume of solution = 20 mL = 20/1000 = 0.02 L

Mole of solute (Ba(NO3)2) =?

Molarity = mole /Volume

0.5 = Mole of Ba(NO3)2 / 0.02

Cross multiply to express in linear form

Mole of Ba(NO3)2 = 0.5 x 0.02

Mole of Ba(NO3)2 = 0.01 mole

Step 3:

Determination of the number of mole of Na2SO4 that reacted.

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

From the balanced equation above,

1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.

Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.

Step 4:

Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:

Mole of Na2SO4 = 0.01 mole

Molarity of Na2SO4 = 0.225M

Volume =?

Molarity = mole /Volume

0.225 = 0.01 / volume

Cross multiply to express in linear form

0.225 x Volume = 0.01

Divide both side by 0.225

Volume = 0.01/0.225

Volume of Na2SO4 = 0.044 L

Converting 0.044 L to mL, we have

Volume of Na2SO4 = 0.044 x 1000

Volume of Na2SO4 = 44 mL

Therefore, 44 mL of Na2SO4 is needed for the reaction

6 0
3 years ago
Read 2 more answers
Which product of petroleum is normally used as fuel in cars and light vehicles​
Elena L [17]

Answer:

mostly gasoline in cars nowadays

8 0
3 years ago
Read 2 more answers
Equal volumes of SO2(g) and O2(g) at STP contain the same number of
LuckyWell [14K]

Answer:

Equal volumes of SO2(g) and O2(g) at STP contain the same number of molecules

Explanation:

According to Avogadro Law,

Equal volume of all the gases at same temperature and pressure have equal number of molecules.

This law state that volume and number of moles of gas have direct relation.

When the amount of gas increases its volume will increase and when the amount of gas decreases its volume will decrease.

Mathematical relation:

V ∝ n

V/n = K

K is proportionality constant.

When number of moles change from n₁ to n₂ and volume from V₁ to V₂

expression will be,

V₁/n₁ = K     ,     V₂/n₂ = K

V₁/n₁ = V₂/n₂

8 0
3 years ago
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You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.
Nitella [24]

The question requires us to use the dilution formula M_iV_i=M_fV_f, where M_i and V_i are the stock concentration and volume respectively, then M_f and V_f are the dilute concentration and volume respectively.

a. C_s_t_o_c_k= 1.5 M KCI, C_d_i_l_u_t_e=0.20M KCl, V_d_i_l_u_t_e=250ml KCl

b.M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.


8 0
3 years ago
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