CH-ch-ch-ch-ch-ch-naoh--------------->ch3-oh
Cl-CH3_CH2_CH2_CH3+C2H5oNa_> C6H15o+
CH3-CH(I)-CH2-CH2-CH2-CH3 + CH3-ONa --------> CH3-CH(O-CH3)-CH2-CH2-CH2-CH3 + NaI
Sodium + Bromine ----> Sodium bromide2 Na + Br2 ----> 2 NaBr
Sodium Hydroxide + Hydrocloric acid --> Sodiumchloride + Water
SN2 and SN1
Cl-CH3_CH2_CH2_CH3+C2H5oNa_> C6H15o+
CH3-CH(I)-CH2-CH2-CH2-CH3 + CH3-ONa --------> CH3-CH(O-CH3)-CH2-CH2-CH2-CH3 + NaI
Sodium + Bromine ----> Sodium bromide2 Na + Br2 ----> 2 NaBr
the equation for sodium acetate with water is NaC2H3O2+2(H2O)=Na+C2H3O2(solid).
This equation is 2 Na + Br2 -> 2 NaBr.
The formula [not equation!] of sodium oxide is Na2O. A possible equation for forming it is 4 Na + O2 -> 2 Na2O.
Methoxide is your nucleophile; it will attack the C3 of 2,2-dimethyloxirane (the carbon that is attached to the oxygen, but doesn't have any methyl groups) via backside attack. This will cause the bond between C3 and the oxygen to break, thus releasing the chain and forming the oxide form of 1-methoxy-2-methyl-2-propanol. Since the oxide of this product is more basic than methanol, it will rip the hydrogen off of methanol, which will create the final product and regenerate methoxide.
Sodium Hydroxide + Hydrocloric acid --> Sodiumchloride + Water
The chemical equation is:2 NaHCO3---------------------Na2O + 2 CO2 + H2O
SN2 and SN1
2 Benzyl alcohol + 2 Na ---> H2(g) + 2 sodium benzoate
the equation for sodium nitrate and copper III iodide can be given below.Cu I 2 +2 Na No3 ->I (NO3)2 + 2NaI. this is the balance reaction for the above.