An LRT has nothing to do with any contrast, but instead is comparing the model fit with and without certain coefficients. So if you do

dds <- DESeq(<deseq object goes here>, "LRT", reduced = ~ compound + time)

Then you are comparing the likelihood for the full model (~ compound + time + compound:time) versus the reduced model (~ compound + time), and if the full model fits the data better, you will get a significant p-value for the likelihood ratio test. In effect what that means is that the time-resolved gene expression is different at one or more time points for at least one of the compounds. This is an omnibus test (similar to an F-test) that tells you things are different without saying what the differences are.

You could come back with Wald tests for each timepoint to nail that down, but that is pretty boring and complex all at the same time. You would probably be better off taking the set of significant genes, using K-means clustering to resolve into multiple patterns and then plot the data for each cluster. You will have to play with K to get reasonable clusters, but that allows you to say things about each cluster, and how the treatments affect the time-dependent resolution of gene expression within each cluster.

Anyway, to show further what I mean about LRT not being contrast-dependent, consider this:

> library(DESeq2)
> z <- makeExampleDESeqDataSet(m = 24)
> colData(z) <- DataFrame(Time = factor(rep(1:4, each = 6)), Treatment = factor(rep(c("A","B"), each = 3, times = 4))) 
> design(z) <- ~ Treatment + Time + Treatment:Time
> dds <- DESeq(z, "LRT", reduced = ~ Treatment + Time)
estimating size factors
estimating dispersions
gene-wise dispersion estimates
mean-dispersion relationship
final dispersion estimates
fitting model and testing              
> resultsNames(dds)
[1] "Intercept"       
[2] "Treatment_B_vs_A"
[3] "Time_2_vs_1"     
[4] "Time_3_vs_1"     
[5] "Time_4_vs_1"     
[6] "TreatmentB.Time2"
[7] "TreatmentB.Time3"
[8] "TreatmentB.Time4"
## let's do a contrast comparing  Time_2_vs_1 vs Time_3_vs_1
## and compare that to the default       
> all.equal(results(dds), results(dds, contrast = c(0,0,1,-1,0,0,0,0)), check.attributes = FALSE)
[1] TRUE

The results with and without a contrast are identical!

Specifying the contrast shouldn't affect the results for an LRT test because it's not a Wald test, so a contrast doesn't make much sense in that context. Which is why I suspect Mike asked you to find somebody local to help you. An alternative, if you need to do the analysis yourself is to educate yourself about linear regression. In the end, you will have to explain what you have done and would presumably want your explanation to make sense, particularly if you are explaining to someone with a statistical background. There's only two ways that can happen - either you know what you are talking about, or you have somebody who does know what she's talking about to provide the explanation. If you can't do the latter, you should probably do the former.